Đáp án:
$\begin{array}{l}
- 2{x^2} + 3x + 1 = - 2\left( {{x^2} - \frac{3}{2}x} \right) + 1\\
= - 2\left( {{x^2} - 2.\frac{3}{4}x + \frac{9}{{16}}} \right) + \frac{{17}}{8}\\
= - 2{\left( {x - \frac{3}{4}} \right)^2} + \frac{{17}}{8} \le \frac{{17}}{8}\forall x\\
\Rightarrow GTLN:y = \frac{{17}}{8} \Leftrightarrow x = \frac{3}{4}\\
a)4{x^2} + {y^2} - 2x + y + 1\\
= \left( {4{x^2} - 2x} \right) + \left( {{y^2} + y} \right) + 1\\
= 4\left( {2{x^2} - 2.\sqrt 2 x.\frac{1}{{2\sqrt 2 }} + \frac{1}{8}} \right) - 4.\frac{1}{8} + {\left( {y + \frac{1}{2}} \right)^2} - \frac{1}{4} + 1\\
= 4{\left( {\sqrt 2 x - \frac{1}{{2\sqrt 2 }}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} + \frac{1}{4} \ge \frac{1}{4}\forall x,y\\
\Rightarrow GTNN:y = \frac{1}{4} \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{1}{4}\\
y = - \frac{1}{2}
\end{array} \right.\\
b)2{x^2} - 2x + 9 - 2xy + {y^2}\\
= {x^2} + {y^2} + 1 - 2x - 2xy + 2y + {x^2} + 8\\
= {\left( { - x + y + 1} \right)^2} + {x^2} + 8 \ge 8\forall x,y\\
\Rightarrow GTNN:y = 8 \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = - 1
\end{array} \right.
\end{array}$
